Update 1:

Below is my original post which I initially deleted, because there were issues with displaying the equations and also I wasn't very confident if it really makes sense. But later, I figured that the optimisation problem that I described can be actually solved efficiently with DP (Dynamic programming).

So I did a sample C++ implementation. Here are some results:

Here is a live demo that you can play with in your browser (make sure browser support WebGL2, like Chrome or Firefox). It takes a bit to load the page.

Here is the C++ implementation: link

Update 2:

Turns out the proposed solution has the following nice property - we can easily control the importance of the two parts F₁ and F₂ of the cost function. Simply change the cost function to F(α)=F₁ + αF₂, where α >= 1.0 is a free parameter. The DP algorithm remains the same.

Here are some result for different α values using the same number of intervals N:

Live demo ^{(WebGL2 required)}

As can be seen, higher α means it is more important to cover the original input intervals even if this means covering more of the background in-between.

Original post

Even-though some good algorithms have already been proposed, I would like to propose a slightly unusual approach - interpreting the task as an optimisation problem. Although, I don't know how to efficiently solve the optimisation problem (or even if it can be solved in reasonable time at all), it might be useful to someone purely as a concept.

First, without loss of generality, lets declare the blue color to be background. We will be painting N green intervals on top of it (N is the number provided to the downsample() function in OP's description). The i^th interval is defined by its starting coordinate 0 <= x_i < x_max and width w_i >= 0 (x_max is the maximum coordinate from the input).

Lets also define the array G(x) to be the number of green cells in the interval [0, x) in the input data. This array can easily be pre-calculated. We will use it to quickly calculate the number of green cells in arbitrary interval [x, y) - namely: G(y) - G(x).

We can now introduce the first part of the cost function for our optimisation problem:

The smaller F₁ is, the better our generated intervals cover the input intervals, so we will be searching for x_i, w_i that minimise it. Ideally we want F₁=0 which would mean that the intervals do not cover any of the background (which of course is not possible because N is less than the input intervals).

However, this function is not enough to describe the problem, because obviously we can minimise it by taking empty intervals: F₁(x, 0)=0. Instead, we want to cover as much as possible from the input intervals. Lets introduce the second part of the cost function which corresponds to this requirement:

The smaller F₂ is, the more input intervals are covered. Ideally we want F₂=0 which would mean that we covered all of the input rectangles. However, minimising F₂ competes with minimising F₁.

Finally, we can state our optimisation problem: find x_i, w_i that minimize F=F₁ + F₂

How to solve this problem? Not sure. Maybe use some metaheuristic approach for global optimisation such as Simulated annealing or Differential evolution. These are typically easy to implement, especially for this simple cost function.

Best case would be to exist some kind of DP algorithm for solving it efficiently, but unlikely.

I would suggest using K-means it is an algorithm used to group data(a more detailed explanation here: https://en.wikipedia.org/wiki/K-means_clustering and here https://scikit-learn.org/stable/modules/generated/sklearn.cluster.KMeans.html) this would be a brief explanation of how the function should look like, hope it is helpful.

from sklearn.cluster import KMeans
import numpy as np

def downsample(input, cluster = 25):

    # you will need to group your labels in a nmpy array as shown bellow
    # for the sake of example I will take just a random array
    X = np.array([[1, 2], [1, 4], [1, 0],[4, 2], [4, 4], [4, 0]])

    # n_clusters will be the same as desired output
    kmeans = KMeans(n_clusters= cluster, random_state=0).fit(X)

    # then you can iterate through labels that was assigned to every entr of your input
    # in our case the interval
    kmeans_list = [None]*cluster
    for i in range(0, X.shape[0]):
        kmeans_list[kmeans.labels_[i]].append(X[i])

    # after that you will basicly have a list of lists and every inner list will contain all points that corespond to a
    # specific label

    ret = [] #return list
    for label_list in kmeans_list:
        left = 10001000 # a big enough number to exced anything that you will get as an input
        right = -left   # same here
        for entry in label_list:
            left = min(left, entry[0])
            right = max(right, entry[1])
        ret.append([left,right])

    return ret

You could do the following:

1. Write out the points that divide the whole strip into intervals as the array [a[0], a[1], a[2], ..., a[n-1]]. In your example, the array would be [0, 5, 6, 10, ... ].
2. Calculate double-interval lengths a[2]-a[0], a[3]-a[1], a[4]-a[2], ..., a[n-1]-a[n-3] and find the least of them. Let it be a[k+2]-a[k]. If there are two or more equal lengths having the lowest value, choose one of them randomly. In your example, you should get the array [6, 5, ... ] and search for the minimum value through it.
3. Swap the intervals (a[k], a[k+1]) and (a[k+1], a[k+2]). Basically, you need to assign a[k+1]=a[k]+a[k+2]-a[k+1] to keep the lengths, and to remove the points a[k] and a[k+2] from the array after that because two pairs of intervals of the same color are now merged into two larger intervals. Thus, the numbers of blue and green intervals decreases by one each after this step.
4. If you're satisfied with the current number of intervals, end the process, otherwise go to the step 1.

You performed the step 2 in order to decrease "color shift" because, at the step 3, the left interval is moved a[k+2]-a[k+1] to the right and the right interval is moved a[k+1]-a[k] to the left. The sum of these distances, a[k+2]-a[k] can be considered a measure of change you're introducing into the whole picture.

Main advantages of this approach:

1. It is simple.
2. It doesn't give a preference to any of the two colors. You don't need to assign one of the colors to be the background and the other to be the painting color. The picture can be considered both as "green-on-blue" and "blue-on-green". This reflects quite common use case when two colors just describe two opposite states (like the bit 0/1, "yes/no" answer) of some process extended in time or in space.
3. It always keeps the balance between colors, i.e. the sum of intervals of each color remains the same during the reduction process. Thus the total brightness of the picture doesn't change. It is important as this total brightness can be considered an "indicator of completeness" at some cases.

I would advise you to use Haar wavelet. That is a very simple algorithm which was often used to provide the functionality of progressive loading for big images on websites.

Here you can see how it works with 2D function. That is what you can use. Alas, the document is in Ukrainian, but code in C++, so readable:)

This document provides an example of 3D object:

Pseudocode on how to compress with Haar wavelet you can find in Wavelets for Computer Graphics: A Primer Part 1y.

Here's another attempt at dynamic programming that's slightly different than Georgi Gerganov's, although the idea to try and formulate a dynamic program may have been inspired by his answer. Neither the implementation nor the concept is guaranteed to be sound but I did include a code sketch with a visual example :)

The search space in this case is not reliant on the total unit width but rather on the number of intervals. It's O(N * n^2) time and O(N * n) space, where N and n are the target and given number of (green) intervals, respectively, because we assume that any newly chosen green interval must be bound by two green intervals (rather than extend arbitrarily into the background).

The idea also utilises the prefix sum idea used to calculate runs with a majority element. We add 1 when we see the target element (in this case green) and subtract 1 for others (that algorithm is also amenable to multiple elements with parallel prefix sum tracking). (I'm not sure that restricting candidate intervals to sections with a majority of the target colour is always warranted but it may be a useful heuristic depending on the desired outcome. It's also adjustable -- we can easily adjust it to check for a different part than 1/2.)

Where Georgi Gerganov's program seeks to minimise, this dynamic program seeks to maximise two ratios. Let h(i, k) represent the best sequence of green intervals up to the ith given interval, utilising k intervals, where each is allowed to stretch back to the left edge of some previous green interval. We speculate that

h(i, k) = max(r + C*r1 + h(i-l, k-1))

where, in the current candidate interval, r is the ratio of green to the length of the stretch, and r1 is the ratio of green to the total given green. r1 is multiplied by an adjustable constant to give more weight to the volume of green covered. l is the length of the stretch.

JavaScript code (for debugging, it includes some extra variables and log lines):

function rnd(n, d=2){
  let m = Math.pow(10,d)
  return Math.round(m*n) / m;
}

function f(A, N, C){
  let ps = [[0,0]];
  let psBG = [0];
  let totalG = 0;
  A.unshift([0,0]);

  for (let i=1; i<A.length; i++){
    let [l,r,c] = A[i];
    
    if (c == 'g'){
      totalG += r - l;
      let prevI = ps[ps.length-1][1];
      let d = l - A[prevI][1];
      let prevS = ps[ps.length-1][0];
      ps.push(
        [prevS - d, i, 'l'],
        [prevS - d + r - l, i, 'r']
      );
      psBG[i] = psBG[i-1];

    } else {
      psBG[i] = psBG[i-1] + r - l;
    }
  }
  //console.log(JSON.stringify(A));
  //console.log('');
  //console.log(JSON.stringify(ps));
  //console.log('');
  //console.log(JSON.stringify(psBG));

  let m = new Array(N + 1);
  m[0] = new Array((ps.length >> 1) + 1);
  for (let i=0; i<m[0].length; i++)
    m[0][i] = [0,0];

  // for each in N
  for (let i=1; i<=N; i++){
    m[i] = new Array((ps.length >> 1) + 1);
    for (let ii=0; ii<m[0].length; ii++)
      m[i][ii] = [0,0];

    // for each interval
    for (let j=i; j<m[0].length; j++){
      m[i][j] = m[i][j-1];

      for (let k=j; k>i-1; k--){
        // our anchors are the right
        // side of each interval, k's are the left
        let jj = 2*j;
        let kk = 2*k - 1;

        // positive means green
        // is a majority
        if (ps[jj][0] - ps[kk][0] > 0){
          let bg = psBG[ps[jj][1]] - psBG[ps[kk][1]];
          let s = A[ps[jj][1]][1] - A[ps[kk][1]][0] - bg;
          let r = s / (bg + s);
          let r1 = C * s / totalG;
          let candidate = r + r1 + m[i-1][j-1][0];
          
          if (candidate > m[i][j][0]){
            m[i][j] = [
              candidate,
              ps[kk][1] + ',' + ps[jj][1],
              bg, s, r, r1,k,m[i-1][j-1][0]
            ];
          }
        }
      }
    }
  }
  /*
  for (row of m)
    console.log(JSON.stringify(
      row.map(l => l.map(x => typeof x != 'number' ? x : rnd(x)))));
  */
  let result = new Array(N);
  let j = m[0].length - 1;
  for (let i=N; i>0; i--){
    let [_,idxs,w,x,y,z,k] = m[i][j];
    let [l,r] = idxs.split(',');
    result[i-1] = [A[l][0], A[r][1], 'g'];
    j = k - 1;
  }

  return result;
}

function show(A, last){
  if (last[1] != A[A.length-1])
    A.push(last);

  let s = '';
  let j;

  for (let i=A.length-1; i>=0; i--){
    let [l, r, c] = A[i];
    let cc = c == 'g' ? 'X' : '.';
    for (let j=r-1; j>=l; j--)
      s = cc + s;
    if (i > 0)
      for (let j=l-1; j>=A[i-1][1]; j--)
        s = '.' + s
  }

  for (let j=A[0][0]-1; j>=0; j--)
    s = '.' + s

  console.log(s);
  return s;
}

function g(A, N, C){
  const ts = f(A, N, C);
  //console.log(JSON.stringify(ts));
  show(A, A[A.length-1]);
  show(ts, A[A.length-1]);
}


var a = [
  [0,5,'b'],
  [5,9,'g'],
  [9,10,'b'],
  [10,15,'g'],
  [15,40,'b'],
  [40,41,'g'],
  [41,43,'b'],
  [43,44,'g'],
  [44,45,'b'],
  [45,46,'g'],
  [46,55,'b'],
  [55,65,'g'],
  [65,100,'b']
];

// (input, N, C)
g(a, 2, 2);
console.log('');
g(a, 3, 2);
console.log('');
g(a, 4, 2);
console.log('');
g(a, 4, 5);